The Stacks project

Lemma 10.12.10. Let

\begin{align*} M_1\xrightarrow {f} M_2\xrightarrow {g} M_3 \to 0 \end{align*}

be an exact sequence of $R$-modules and homomorphisms, and let $N$ be any $R$-module. Then the sequence

10.12.10.1
\begin{equation} \label{algebra-equation-2ndex} M_1\otimes N\xrightarrow {f \otimes 1} M_2\otimes N \xrightarrow {g \otimes 1} M_3\otimes N \to 0 \end{equation}

is exact. In other words, the functor $- \otimes _ R N$ is right exact, in the sense that tensoring each term in the original right exact sequence preserves the exactness.

Proof. For every $R$-module $P$ we apply the functor $\mathop{\mathrm{Hom}}\nolimits (-, \mathop{\mathrm{Hom}}\nolimits (N, P))$ to the first exact sequence. We obtain

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits (M_3, \mathop{\mathrm{Hom}}\nolimits (N, P)) \to \mathop{\mathrm{Hom}}\nolimits (M_2, \mathop{\mathrm{Hom}}\nolimits (N, P)) \to \mathop{\mathrm{Hom}}\nolimits (M_1, \mathop{\mathrm{Hom}}\nolimits (N, P)) \]

which is exact by Lemma 10.10.1 (1). By Lemma 10.12.8 this becomes the sequence

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits (M_3 \otimes N, P) \to \mathop{\mathrm{Hom}}\nolimits (M_2 \otimes N, P) \to \mathop{\mathrm{Hom}}\nolimits (M_1 \otimes N, P) \]

which is therefore also exact. Then using Lemma 10.10.1 (1) again, we arrive at the desired exact sequence. $\square$

Comments (2)

Comment #10136 by on

A few minor suggestions: 1) Start the first sentence with "For every -module , ". 2) Say that the first displayed sequence is exact by tag 0582, part (1). 3) Instead of writing "Using the pullback property again", refer to tag 0582, part (1) explicitly.

There are also:

  • 8 comment(s) on Section 10.12: Tensor products

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